Find the Taylor series for f(x,y)=\log(4x-3y), up to and including the quadratic terms about the point P=(1,1)

Can someone check my working:

f_x=\frac{4}{4x-3y}\Rightarrow f_x(1,1)=4
f_y=\frac{-3}{4x-3y}\Rightarrow f_y(1,1)=-3
f_{xx}=\frac{-16}{(4x-3y)^2}\Rightarrow f_x(1,1)=-16
f_{yy}=\frac{-9}{(4x-3y)^2}\Rightarrow f_x(1,1)=-9
f_{xy}=\frac{12}{(4x-3y)^2}\Rightarrow f_x(1,1)=12

\ln(4x-3y)\approx 4(x-1)-3(y-1)+\frac{1}{2!}(-16(x-1)^2+24(x-1)(y-1)-9(y-1)^2)

2) Using the constant and linear terms of the Taylor series above, find a first order approximation to f(P+(0.01,-0.01))

So for this would I just use

\ln(4x-3y)\approx 4(x-1)-3(y-1)+12(x-1)(y-1)

and substitute (1.01,0.99)